91Ó°ÊÓ

An electron in an external magnetic field ${\stackrel{â‡\P Ä}{B}}_{ext}$ has its spin angular momentum S_z antiparallel to ${\stackrel{â‡\P Ä}{B}}_{ext}$.

Using the formula for the potential for the electron and inserting the spin angular magnetic momentum, it can be predictedwhether energy should be supplied to or lost bytheelectron if the electron undergoes a spin-flop.

The formulae are as follows:

$\begin{array}{rcl}U& =& -\stackrel{â‡\P Ä}{{\mathrm{μ}}_s}.{\stackrel{â‡\P Ä}{B}}_{ext}.\\ {\mathrm{μ}}_{s,z}& =& -\frac{e}{m}{S}_z.\\ S_z& =& m_s\frac{h}{2\mathrm{Ï€}}.\end{array}$

The potential energy for the electron is given by,

$$\begin{gathered} U=-{\mathrm{μ}}_{s,z}.B_{ext,} \\ And, \\ {\mathrm{μ}}_{s,z}=-\frac{e}{m}{S}_z, \end{gathered}$$

But,

$S_z=m_s\frac{h}{2\mathrm{Ï€}},$

When the spin angular momentum S_z is parallel to ${\stackrel{â‡\P Ä}{B}}_{ext}$, then role="math" localid="1663044479616" $m_s=+\frac{1}{2}$, and when it is antiparallel ${\stackrel{â‡\P Ä}{B}}_{ext}$to, $m_s=-\frac{1}{2}$.

Hence, the potential energy of the electron when its spin angular momentum S_z is parallel to${\stackrel{â‡\P Ä}{B}}_{ext}$after spin-flop is,

$U=\frac{e}{m}{S}_z.B_{ext},$

This implies that energy must be supplied totheelectron if the electron undergoes a spin-flop.

Whether energy is supplied to or lost by an electron if the electron undergoes a spin-flop depends on the sign of the spin magnetic quantum number.